HYACINTHOS 23086

Antreas Hatzipolakis

 

 

[APH]:

Let ABC be a triangle and Na,Nb,Nc the NPC centers of
IBC,ICA, IAB, resp.

Denote:

Ka, Kb, Kc = the isogoanal conjugates of Na,Nb,Nc
wrt IBC,ICA,IAB, resp.

ie Ka, Kb, Kc = the Kosnita points X(54) of
IBC,ICA,IAB, resp.

Ka,Kb,Kc lie on the OI line.

The circumcircles of AKbKc, BKcKa, CKaKb, ABC
are concurrent.

Which is the radical center of the circumcircles of
KaBC, KbCA, KcAB?



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[César Lozada]

Antreas:

 

>The circumcircles of AKbKc, BKcKa, CKaKb, ABC are concurrent.

 

 

Answer:

Z= a/((b-c)*(b^4+a*b^3+4*b^2*c*a-2*b^2*c^2-3*a^2*b^2+4*b*c^2*a-4*b*c*a^2-a^3*b+c^3*a-c*a^3+2*a^4-3*c^2*a^2+c^4)) : : (trilinears)
ETC_SEARCH: -0.415179481263028, 0.86263677436611, 3.235075706390779

Z lie on circumcircle and  line (X(1385)X(2687))

 

 

>Which is the radical center of the circumcircles of KaBC, KbCA, KcAB?

 

Answer: X(484)=1^st Evans perspector


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[Angel Montesdeoca]:


*** Kosnita points of IBC

Ka = (a^2(a-b)(a-c) : b(a-b)(a^2- b^2+c^2-a(b+2c)) : c(a-c)(a^2+b^2-c^2-a(2b+c))

*** The circumcircles of AKbKc, BKcKa, CKaKb, ABC are concurrent in

( a^2/((b-c)(2a^4-a^3(b+c)-a^2(

3b^2+4b*c+3c^2)+a(b+c)(b^2+3b*c+c^2)+(b^2-c^2)^2)) : ... : ... ),

with (6-9-13)-search number -0.4151794812633478152167

*** The circumcircles of KaBC, KbCA, KcAB are concurrent in X(484).